|
Question 435012: Need to see if this is a valid question:
Safe working load model for wire rope is: 4*D^2 = S
S is in Tons
D is in Inches
A 9-ton load requires a 1.5 inch diameter wire to lift it.
Q: When determining the safe working load S of a rope that is old or worn, decrease S by 50%. Write a model for S when using an old wire rope. What diameter of old wire rope do you need to safely lift a 9-ton load.
Why Decrease the Working Load by 50%? How do I solve this or is it valid? My equation results in a 1.06 inch Diameter. That has to be WRONG.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Need to see if this is a valid question:
Safe working load model for wire rope is: 4*D^2 = S
S is in Tons
D is in Inches
A 9-ton load requires a 1.5 inch diameter wire to lift it.
--------------------------------
Q: When determining the safe working load S of a rope that is old or worn, decrease S by 50%. Write a model for S when using an old wire rope.
---
Yes, that is confusing.
A worn rope of 1.5 in will only handle 4.5 tons
----
You had D^2= S/4 for a new rope
-----
Model for worn rope = ?:
You want a 1.5 in rope to only handle 9/2 tones
Solve k(3/2)^2 = 9/2
k(9/4) = 9/2
k = (4/9)(9/2)
k = 2
----
Model for worn rope:
2D^2 = S
----
What diameter of old wire rope do you need to safely lift a 9-ton load.
Solve: 2D^2= 9 tons
D^2 = 9/2
D = 3/sqrt(2)
D = 2.1213 inches
=========================
Why Decrease the Working Load by 50%?
The worn rope cannot safely handle as much weight
as a new rope. It can only handle half as much
weight as a new rope.
------------------
Cheers,
Stan H.
=================
|
|
|
| |
